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In this video, we’re going to be talking about the bulk properties of an ideal gas.
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In particular, we’ll be looking at a very useful equation, known as the ideal gas equation, which links together three important properties of a gas — volume, temperature, and pressure.
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But to start off with, let’s define exactly what’s meant by the bulk properties of an ideal gas.
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Firstly, what do we mean by bulk properties?
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Well, bulk properties describe the average or collective behavior of a large number of particles.
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We know that gases consist of gas molecules.
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And even a small amount of gas contains a large number of molecules.
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This means that in most situations, if we tried to describe the individual properties of each molecule in a gas, for example, their velocities and their positions, we can see that things would get incredibly complicated very quickly.
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Especially considering that we’d have to measure these quantities in three dimensions and that constantly changing.
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So instead of all this, it’s much easier and more useful to just consider the bulk properties of a gas.
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In the case of an ideal gas, the bulk properties we’re interested in are the volume, pressure, and temperature of the gas.
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These properties group the behavior of all of the molecules in a gas together.
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So we can think of a gas as a single cohesive thing, rather than as billions of individual molecules.
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The second thing we need to define is what we mean by an ideal gas.
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When we use the word ideal in day-to-day life, we use it to describe something which is really good or convenient.
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When we use the word ideal in science, it has a similar meaning.
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We use it to refer to a convenient, simplified description of something.
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So when we talk about an ideal gas, we’re talking about a simplified approximation of how our gas behaves.
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Specifically, we define an ideal gas as consisting of particles which don’t interact with each other and which have negligible size.
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In other words, we’re simplifying our description of a gas by saying that there are no forces between the gas molecules such as electrostatic repulsion and that the volume of the gas molecules is negligible compared to the volume of the entire gas.
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The reason that we simplify our description in this way is that it makes the maths we need to describe that gas much simpler while still being accurate enough to be useful in many situations.
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Physicists have used these two simplifying assumptions to come up with a simple and useful equation that relates these three bulk properties of a gas.
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This equation is known as the ideal gas equation, and it looks like this.
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The ideal gas equation tells us that the pressure of an ideal gas multiplied by its volume is equal to a constant of proportionality multiplied by the temperature of the gas.
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Now, pressure is equivalent to force divided by area.
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When we talk about the pressure of a gas in a container, we’re talking about the pressure that the gas molecules exert on the walls of the container by colliding with them.
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The pressure of a gas is equal to the total force that the gas molecules exert on the inside of that container divided by the total internal area of the walls of the container.
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Because all actions have an equal and opposite reaction, we can also think of this the other way around.
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The pressure that the gas exerts on its container is exactly the same as the pressure that the container exerts on the gas.
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Depending on the context that we’re looking at, it might sometimes be easier to think of the pressure of a gas as being an outward, expansive pressure that acts on its container.
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But at other times, it might be easier to think of pressure as being an inward squeezing pressure that the container exerts on a gas.
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But thinking about it in these different ways doesn’t change how we calculate it or its value.
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When we think about the pressure of a gas in a container, which will tend to be a reasonably small amount of gas, the pressure can be measured at any point throughout the gas.
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And it takes the same values at each of these points.
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The units of pressure are pascals represented by a capital “𝑃” and a lowercase “𝑎.”
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And because pressure is equal to force divided by area, one pascal is equivalent to one newton per meter squared.
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The next variable in this equation, volume, is much easier to conceptualize.
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The volume of a gas is just the amount of 3D space that it occupies, which we measure in meters cubed.
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The other variable in this equation is temperature.
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Like with most equations in physics that involve temperature, it’s really important that the value that we use here is measured in kelvin rather than Celsius.
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We can recall that in the Celsius temperature scale, zero degrees Celsius is arbitrarily defined as being the freezing point of water.
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However, the lowest temperature that an object can theoretically have is actually negative 273.15 degrees Celsius.
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So this temperature, known as absolute zero, is where we define the zero point of the kelvin scale.
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Now, other than the positions of the zero points, the kelvin scale and the degrees Celsius scale are actually the same.
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That is, a temperature increase of one degree Celsius is exactly the same as a temperature increase of one kelvin, which means that the freezing point of water is 273.15 kelvin.
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In other words, the temperature measured in kelvin, which we could call 𝑇 𝑘, is equal to the temperature measured in Celsius, which we could call 𝑇 𝑐 plus 273.15, although we’ll often round this to three significant figures, in which case it’s 273.
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Whenever we use the ideal gas equation, we need to make sure that any temperatures given in Celsius are first converted into kelvin.
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Otherwise, our calculation will give us an incorrect result.
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The final part of this equation to look at is the constant of proportionality 𝑘.
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This constant, like most constants in physical equations, just make sure that when we use the ideal gas equation, we get an answer which is in the correct units.
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For example, if we substituted in a pressure in pascals and a volume in meters cubed, then we could obtain a temperature in kelvin.
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One important thing to know about this constant of proportionality is that it depends on the amount of gas that we have.
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In fact, its value is proportional to the number of gas molecules.
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This means that this constant of proportionality is only a constant if the number of gas molecules is kept the same.
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So it’s important to be aware of this whenever we use the ideal gas equation.
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As long as the number of gas molecules doesn’t change, the ideal gas equation tells us that the pressure of an ideal gas multiplied by its volume is proportional to its temperature.
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So now that we’ve talked about what an ideal gas is and how its bulk properties can be related to each other using the ideal gas equation, let’s take a look at some example questions.
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Which of the following formulas most correctly represents the relation between the pressure, volume, and absolute temperature of an ideal gas?
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(A) 𝑃 over 𝑉 is proportional to 𝑇, (B) 𝑃 over 𝑇 is proportional to 𝑉, (C) 𝑃 times 𝑉 is proportional to 𝑇, (D) 𝑉 over 𝑇 is proportional to 𝑃, or (E) 𝑉 over 𝑃 is proportional to 𝑇.
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So in this question, we’ve been given five different formulas.
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And we need to decide which one of them most accurately represents the relation between these three quantities.
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We can start by recalling which symbols we generally use to represent these quantities.
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Well, we usually use capital 𝑃 for pressure, capital 𝑉 for volume, and a capital 𝑇 for absolute temperature.
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Since each of our available answers uses all three of these symbols, that means that they all represent some kind of relation between the three quantities that we’re interested in.
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The key phrase in our question which can help us figure out which one of these options is most correct is ideal gas.
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We can recall that an ideal gas is a simplified description of a gas based on the assumptions that gas molecules have negligible size and don’t interact with each other.
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Now there’s one key equation that we should always think of whenever we hear the phrase ideal gas.
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And that is the ideal gas equation, 𝑃𝑉 equals 𝑘𝑇.
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This equation gives us the relationship between three of the main bulk properties of an ideal gas — the pressure 𝑃, the volume 𝑉, and the absolute temperature 𝑇— which are, of course, the same as the three quantities we’re being asked about in this question.
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But unfortunately, while this does give us a correct relationship between the pressure, volume, and absolute temperature of an ideal gas, none of our available answers look the same as this equation.
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In fact, one of the first things we might have noticed about the answer options we’ve been given is that while they are formulas, none of them are equations.
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That is, they don’t have equal signs.
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Instead, each of the formulas we’ve been given has this symbol.
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This is the proportionality sign.
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We use it to represent the fact that both sides of the formula are proportional to each other.
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So, for example, if two variables 𝐴 and 𝐵 were proportional to each other, that could be represented by this formula.
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What this means is that while the two variables are not necessarily equal to each other, they do increase or decrease proportionally to each other.
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Which means that, for example, if we were to double the size of 𝐴, that is, multiply it by two, then we’d also find that the size of 𝐵 would double as well.
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Or, for example, if we multiplied 𝐵 by 0.25, then we’d find that 𝐴 is multiplied by 0.25 as well.
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You might recall that it’s possible to turn any proportionality statement into an equality statement.
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That is, we can turn a statement like this involving a proportionality sign into an equation with an equal sign.
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We can do this by simply replacing the proportionality sign with an equal sign and introducing a constant of proportionality, generally represented by a 𝐾, which is multiplied by one of the variables.
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This gives us an equation which expresses exactly the same relationship between 𝐴 and 𝐵, as this proportionality statement does.
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We can see that in this equation, if we were to multiply 𝐵 by two, for example, then because 𝐴 is equal to some number times 𝐵, 𝐴 would also have to increase by two times.
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In this case, the constant of proportionality 𝐾 plays the role of scaling 𝐵 such that 𝐾 times 𝐵 is exactly equal to 𝐴.
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In many physical equations, we also find that the constant of proportionality plays an additional role of making sure that the units on the left-hand side of the equation are equivalent to the units on the right-hand side of the equation.
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It’s also worth noting that there are two ways that we can turn this proportionality statement into an equation.
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We can write the constant of proportionality on the right-hand side of the equation like this such that it’s multiplied by 𝐵.
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Or we could write our equation like this, with the constant of proportionality multiplied by 𝐴.
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It’s really important to note that these values of 𝐾 are not the same because we have effectively defined 𝐾 in a different way in each equation.
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In fact, they’ll even have different units.
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They’re different constants entirely.
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So we should really call one of them 𝐾 one and the other one 𝐾 two to distinguish between them.
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So now, both of these equations and this statement of proportionality are exactly equivalent to each other.
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It’s also important to note that statements of proportionality can have more than one variable on each side of the proportionality sign, which we can see is the case in all of the answer options we’ve been given.
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For example, we could say that 𝐴 is proportional to 𝐵 times 𝐶.
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In this case, if we were to multiply the size of 𝐴 by two, then we would find that the product of 𝐵 and 𝐶 is multiplied by two as well.
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Having more than one variable on either side of a proportionality statement doesn’t change the way that we can turn it into an equation.
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So the statement 𝐴 is proportional to 𝐵 times 𝐶 is equivalent to 𝐴 equals 𝐾 one times 𝐵𝐶 or 𝐾 two times 𝐴 equals 𝐵𝐶.
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Looking back at the question, we can see that the ideal gas equation gives us the relation that we’re looking for.
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However, the available answers are all proportionality statements.
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This means that we can find our answer by turning this equation into a proportionality statement.
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In the ideal gas equation, we have two variables, 𝑃 and 𝑉, on one side of the equation and one variable, 𝑇, on the other.
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The presence of this constant of proportionality means the equation tells us 𝑃 times 𝑉 is proportional to 𝑇.
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These two statements are equivalent ways of expressing the same relation between the three variables.
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And we can also see that this statement of proportionality is one of our answer options.
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So option (C) is the correct answer.
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The formula that most correctly represents the relation between the pressure, volume, and absolute temperature of an ideal gas is 𝑃 times 𝑉 is proportional to 𝑇.
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Now that we’ve answered that, let’s look at another example question.
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A gas cylinder holds 3.25 meters cubed of gas at a pressure of 520 kilopascals and a temperature of 300 kelvin.
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At what temperature would the gas pressure in the cylinder become 865 kilopascals?
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Answer to three significant figures.
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So this question describes a cylinder of gas with a volume of 3.25 meters cubed, a pressure of 520 kilopascals, which is the same as 520,000 pascals, and a temperature of 300 kelvin.
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We’re being asked to calculate the temperature at which the gas pressure in the cylinder would become 865 kilopascals or 865,000 pascals.
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So we could imagine the same cylinder of gas at sometime later where the pressure has changed to 865 kilopascals or 865,000 pascals and the temperature has taken some unknown value that we need to work out.
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And because it’s the same cylinder, we can assume that the volume would go unchanged.
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At this point, because we have two different sets of values for the quantities volume, pressure, and temperature, let’s call the initial set of values on the left 𝑉 one, 𝑃 one, and 𝑇 one and the changed values on the right can be 𝑉 two, 𝑃 two, and 𝑇 two.
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So our challenge in this question is to find 𝑇 two.
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So to start with, let’s think of the variables that we have to deal with in this question and see if we can think of any equations that might help us.
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Well, this question asks us to consider the volume, pressure, and temperature of the gas.
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One equation that gives us the relationship between these quantities is the ideal gas equation.
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This tells us that the pressure of an ideal gas multiplied by its volume is equal to some constant multiplied by its absolute temperature.
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We can recall that the ideal gas equation makes the assumptions that gas molecules have negligible size and don’t interact with each other.
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Even though this question talks about a real gas, it’s still reasonable to use the ideal gas equation as it can still give us accurate answers without being too complicated.
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Now, in this question, we’re trying to find the value of 𝑇.
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So let’s rearrange the ideal gas equation to make 𝑇 the subject and do this by just dividing both sides of the ideal gas equation by 𝑘, giving us 𝑃𝑉 over 𝑘 equals 𝑇.
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Since we need to find the temperature of the gas after the pressure has increased, it seems reasonable that we can just plug these values, 𝑉 two and 𝑃 two, into this equation.
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And it will tell us 𝑇 two.
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However, while this is technically true, if we tried doing this, we’ll quickly realize that we don’t know what the value of 𝑘 is.
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This is because 𝑘 actually takes different values depending on the number of gas molecules that we’re dealing with.
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So it’s not something we can just look up.
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So as it stands, we can’t substitute 𝑉 two and 𝑃 two straight into the equation and just get a value of 𝑇 two.
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However, because we have a set of initial conditions for the volume, pressure, and temperature of the gas, there is something else we can do with the ideal gas equation to find the answer.
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If we start with 𝑃𝑉 equals 𝑘𝑇 and divide both sides of the equation by 𝑇, we get 𝑃𝑉 over 𝑇 equals 𝑘.
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Because the amount of gas molecules in this question is constant, this means that 𝑘 is constant.
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So regardless of how we try to change the pressure, volume, or temperature of a fixed amount of gas, we find that the value of 𝑃𝑉 over 𝑇 is always constant.
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In other words, 𝑃 one times 𝑉 one divided by 𝑇 one is equal to 𝑃 two times 𝑉 two divided by 𝑇 two.
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Writing the ideal gas equation in this way is actually equivalent to its more familiar form 𝑃𝑉 equals 𝑘𝑇.
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However, it makes it easy to calculate how variables change without needing to know 𝑘.
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And we can use this formulation for this question.
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Since we’re trying to find 𝑇 two, let’s first rearrange this to make 𝑇 two the subject.
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First, we can multiply both sides of the equation by 𝑇 two, then multiply both sides of the equation by 𝑇 one, and finally divide both sides of the equation by 𝑃 one 𝑉 one.
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With the equation in this form because we know the initial temperature, initial pressure, and initial volume before the pressure increased and we know the volume and the pressure after the pressure increased.
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We can just substitute all of these values in and calculate 𝑇 two.
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Let’s just give ourselves some more space.
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And we know that 𝑇 one is 300 kelvin, 𝑃 two is 865,000 pascals, 𝑉 two is 3.25 meters cubed, 𝑃 one is 520,000 pascals, and 𝑉 one is 3.25 meters cubed as well.
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Plugging all this into our calculator, we get an answer of 499.04 which has units of kelvin since it’s a temperature.
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And rounding our answer to three significant figures gives us a final answer of 499 kelvin.
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Now that we’ve looked at some example questions, let’s summarize what we’ve talked about in this video.
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Firstly, we defined an ideal gas by the assumptions that particles in a gas have negligible size and don’t interact with each other.
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We also defined bulk properties as properties that result from the average behavior of many particles.
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And the bulk properties of an ideal gas are pressure, volume, and temperature.
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We’ve seen how these three quantities are related by the ideal gas equation, 𝑃𝑉 equals 𝑘𝑇, where 𝑘 is proportional to the number of gas molecules.
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And finally, we’ve seen how we can use the ideal gas equation to obtain the formula 𝑃 one 𝑉 one over 𝑇 one equals 𝑃 two 𝑉 two over 𝑇 two.
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And we’ve shown how we can use this equation to calculate changes in the pressure, volume, and temperature of the gas without needing to know the value of 𝑘.